A total of 78 seats for a local concert are sold producing a total revenue of $483. If seats cost either $2.50 or $10.50, how many of the $10.50 seats were sold?

Answer 1

Let's define two variables:

• Let \( x \) be the number of $2.50 seats sold.
• Let \( y \) be the number of $10.50 seats sold.

We have the following information from the problem:

1. The total number of seats sold is 78: \[ x + y = 78 \]

2. The total revenue from the seats is $483: \[ 2.50x + 10.50y = 483 \]

Now we can solve this system of equations.

First, let's solve the first equation for \( x \): \[ x = 78 - y \]

Next, we substitute this expression for \( x \) into the second equation: \[ 2.50(78 - y) + 10.50y = 483 \] Distributing \( 2.50 \): \[ 2.50 \cdot 78 - 2.50y + 10.50y = 483 \] Calculating \( 2.50 \cdot 78 \): \[ 195 - 2.50y + 10.50y = 483 \] Combining like terms: \[ 195 + 8y = 483 \] Now, we isolate \( y \): \[ 8y = 483 - 195 \] \[ 8y = 288 \] \[ y = \frac{288}{8} = 36 \]

Now that we've found \( y \), which represents the number of $10.50 seats sold, we get: \[ \boxed{36} \]

To find \( x \), the number of $2.50 seats sold: \[ x = 78 - y = 78 - 36 = 42 \]

Now, let’s verify the revenue: \[ 2.50x + 10.50y = 2.50(42) + 10.50(36) \] Calculating: \[ 2.50 \times 42 = 105 \] \[ 10.50 \times 36 = 378 \] Adding these amounts: \[ 105 + 378 = 483 \]

Thus, everything checks out. The number of $10.50 seats sold is indeed \( \boxed{36} \).